Prove by induction that the integer next greater than `(3+sqrt(5))^n` is divisible by `2^n` for all `n in N`.

Explanation

Let `alpha=3+sqrt(5) and beta =3-sqrt(5)`
`therefore 0 lt beta^(n)lt 1,forall n in N`
`rArr alpha +beta=6,alpha beta=4` .....(i)
Then , `alpha and beta` are the roots of `x^2-6x+4=0`
`rArr alpha^2=6alpha-4`
`beta^2=6beta-4` ...(ii)
`therefore alpha^2+beta^2=6(alpha+beta)-8=28` ......(iii)
`therefore alpha^n+beta^n=(3+sqrt(5))^n+(3-sqrt(5))^n`
`=2[3^n+.^(n)C_(2)3^(n-2).5+.^(n)C_(4)3^(n-4).5^(2)+........]`
Even integer .
As , `0lt beta^(n) lt 1,alpha^(n)+beta^(n)` is the even integer next greater than `alpha^(n)`.
Step II For `n=1`,
`alpha+beta=6`
divisible by `2^1`
and `n=2, alpha^2+beta^2=28`
divisible by `2^2`
which is true for `n=1,2`.
Step II Assume it is true for `n=k` .
i.e., `alpha^k+beta^k` is divisible by `2^k`.
Step III For `n=k+1`.
the integer next greater than `alpha^(k+1) is alpha^(k+1)+beta^(k+1)`
`= alpha^2.alpha^(k-1)+beta^2.beta^(k-1)`
`=(6alpha-4).alpha^(k-1)+(6beta-4).beta^(k-1)`
`=6(alpha^k+beta^k)-4(alpha^(k-1)+beta_(k-1))`
`=3` (divisible by `2^(k+1))-` (divisible by `2^k+1`)
=Divisible by `2^k+1`.
This shows that the result is true for `n=k+1`. Hence , the integer next greater than `alpha^(k+1)` is divisible by `2^(k+1)`.